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3.5 Q-9

Question Statement

Evaluate the integral:

∫3x2βˆ’12x+11(xβˆ’1)(xβˆ’2)(xβˆ’3),dx\int \frac{3x^{2} - 12x + 11}{(x-1)(x-2)(x-3)} , dx

Background and Explanation

This problem involves the method of partial fractions, which is used to decompose a rational function into simpler fractions that can be integrated easily. Here, the denominator is already factored, and the numerator is of lower degree than the denominator, making the function suitable for partial fraction decomposition.


Solution

We aim to decompose the given fraction:

3x2βˆ’12x+11(xβˆ’1)(xβˆ’2)(xβˆ’3).\frac{3x^{2} - 12x + 11}{(x-1)(x-2)(x-3)}.

Step 1: Setup Partial Fraction Form

We express the fraction as:

3x2βˆ’12x+11(xβˆ’1)(xβˆ’2)(xβˆ’3)=Axβˆ’1+Bxβˆ’2+Cxβˆ’3.\frac{3x^{2} - 12x + 11}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}.

Step 2: Multiply to Eliminate Denominators

Multiply through by the denominator (xβˆ’1)(xβˆ’2)(xβˆ’3)(x-1)(x-2)(x-3):

3x2βˆ’12x+11=A(xβˆ’2)(xβˆ’3)+B(xβˆ’1)(xβˆ’3)+C(xβˆ’1)(xβˆ’2).3x^{2} - 12x + 11 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2).

Step 3: Solve for Constants AA, BB, and CC

  1. Let x=1x = 1:
3(1)2βˆ’12(1)+11=A(1βˆ’2)(1βˆ’3), 3(1)^2 - 12(1) + 11 = A(1-2)(1-3), 3βˆ’12+11=A(βˆ’1)(βˆ’2), 3 - 12 + 11 = A(-1)(-2), 2=2Aβ€…β€ŠβŸΉβ€…β€ŠA=1. 2 = 2A \implies A = 1.
  1. Let x=2x = 2:
3(2)2βˆ’12(2)+11=B(2βˆ’1)(2βˆ’3), 3(2)^2 - 12(2) + 11 = B(2-1)(2-3), 3(4)βˆ’24+11=B(1)(βˆ’1), 3(4) - 24 + 11 = B(1)(-1), 12βˆ’24+11=βˆ’Bβ€…β€ŠβŸΉβ€…β€Šβˆ’1=βˆ’Bβ€…β€ŠβŸΉβ€…β€ŠB=1. 12 - 24 + 11 = -B \implies -1 = -B \implies B = 1.
  1. Let x=3x = 3:
3(3)2βˆ’12(3)+11=C(3βˆ’1)(3βˆ’2), 3(3)^2 - 12(3) + 11 = C(3-1)(3-2), 3(9)βˆ’36+11=C(2)(1), 3(9) - 36 + 11 = C(2)(1), 27βˆ’36+11=2Cβ€…β€ŠβŸΉβ€…β€Š2=2Cβ€…β€ŠβŸΉβ€…β€ŠC=1. 27 - 36 + 11 = 2C \implies 2 = 2C \implies C = 1.

Step 4: Rewrite the Integral

Substitute AA, BB, and CC into the partial fraction decomposition:

3x2βˆ’12x+11(xβˆ’1)(xβˆ’2)(xβˆ’3)=1xβˆ’1+1xβˆ’2+1xβˆ’3.\frac{3x^{2} - 12x + 11}{(x-1)(x-2)(x-3)} = \frac{1}{x-1} + \frac{1}{x-2} + \frac{1}{x-3}.

The integral becomes:

∫3x2βˆ’12x+11(xβˆ’1)(xβˆ’2)(xβˆ’3),dx=∫1xβˆ’1,dx+∫1xβˆ’2,dx+∫1xβˆ’3,dx.\int \frac{3x^{2} - 12x + 11}{(x-1)(x-2)(x-3)} , dx = \int \frac{1}{x-1} , dx + \int \frac{1}{x-2} , dx + \int \frac{1}{x-3} , dx.

Step 5: Evaluate Each Term

  1. For ∫1xβˆ’1,dx\int \frac{1}{x-1} , dx:
ln⁑∣xβˆ’1∣. \ln |x-1|.
  1. For ∫1xβˆ’2,dx\int \frac{1}{x-2} , dx:
ln⁑∣xβˆ’2∣. \ln |x-2|.
  1. For ∫1xβˆ’3,dx\int \frac{1}{x-3} , dx:
ln⁑∣xβˆ’3∣. \ln |x-3|.

Combine results:

∫3x2βˆ’12x+11(xβˆ’1)(xβˆ’2)(xβˆ’3),dx=ln⁑∣xβˆ’1∣+ln⁑∣xβˆ’2∣+ln⁑∣xβˆ’3∣+C,\int \frac{3x^{2} - 12x + 11}{(x-1)(x-2)(x-3)} , dx = \ln |x-1| + \ln |x-2| + \ln |x-3| + C,

where CC is the constant of integration.


Key Formulas or Methods Used

  1. Partial Fraction Decomposition:
    P(x)Q(x)=Axβˆ’1+Bxβˆ’2+Cxβˆ’3,\frac{P(x)}{Q(x)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3},
    for distinct linear factors in the denominator.

  2. Basic Logarithmic Integrals:
    ∫1xβˆ’a,dx=ln⁑∣xβˆ’a∣+C.\int \frac{1}{x-a} , dx = \ln |x-a| + C.


Summary of Steps

  1. Decompose the fraction into partial fractions:
    3x2βˆ’12x+11(xβˆ’1)(xβˆ’2)(xβˆ’3)=1xβˆ’1+1xβˆ’2+1xβˆ’3.\frac{3x^{2} - 12x + 11}{(x-1)(x-2)(x-3)} = \frac{1}{x-1} + \frac{1}{x-2} + \frac{1}{x-3}.

  2. Integrate each term individually:
    ∫1xβˆ’1,dx,∫1xβˆ’2,dx,∫1xβˆ’3,dx.\int \frac{1}{x-1} , dx, \quad \int \frac{1}{x-2} , dx, \quad \int \frac{1}{x-3} , dx.

  3. Combine results to obtain the final solution:
    ln⁑∣xβˆ’1∣+ln⁑∣xβˆ’2∣+ln⁑∣xβˆ’3∣+C.\ln |x-1| + \ln |x-2| + \ln |x-3| + C.